TABLEAU POSETS AND THE FAKE DEGREES OFkonvalinka/zeros.pdf · The major index statistic on permutations was introduced by Percy MacMahon in his seminal works [Mac13, Mac17]. At rst

TABLEAU POSETS AND THE FAKE DEGREES OFCOINVARIANT ALGEBRAS

SARA C. BILLEY, MATJAŽ KONVALINKA, JOSHUA P. SWANSON

Abstract. We introduce two new partial orders on the standard Youngtableaux of a given partition shape, in analogy with the strong and weakBruhat orders on permutations. Both posets are ranked by the majorindex statistic offset by a fixed shift. The existence of such ranked posetstructures allows us to classify the realizable major index statistics onstandard tableaux of arbitrary straight shape and certain skew shapes.By a theorem of Lusztig–Stanley, this classification can be interpreted asdetermining which irreducible representations of the symmetric group existin which homogeneous components of the corresponding coinvariant algebra,strengthening a recent result of the third author for the modular majorindex. Our approach is to identify patterns in standard tableaux that allowone to mutate descent sets in a controlled manner. By work of Lusztig andStembridge, the arguments extend to a classification of all nonzero fakedegrees of coinvariant algebras for finite complex reflection groups in theinfinite family of Shephard–Todd groups.

1. Introduction

Let SYT(λ) denote the set of all standard Young tableaux of partition shapeλ. We say i is a descent in a standard tableau T if i + 1 comes before i inthe row reading word of T , read from bottom to top along rows in Englishnotation. Equivalently, i is a descent in T if i + 1 appears in a lower row inT . Let maj(T ) denote the major index statistic on SYT(λ), which is definedto be the sum of the descents of T . The major index generating function forSYT(λ) is given by

(1) SYT(λ)maj(q) ∶= ∑T ∈SYT(λ)

qmaj(T ) =∑k≥0

bλ,kqk.

The polynomial SYT(λ)maj(q) has two elegant closed forms, one due toSteinberg based on dimensions of irreducible representations of GLn(Fq), see

Date: May 7, 2020.The first author was partially supported by the Washington Research Foundation and DMS-1764012. The second author was partially supported by Research Project BI-US/16-17-042of the Slovenian Research Agency and research core funding No. P1-0294.

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2 SARA C. BILLEY, MATJAŽ KONVALINKA, JOSHUA P. SWANSON

[Ste51], and one due to Stanley [Sta79] generalizing the Hook-Length Formula;see Theorem 2.11.

For fixed λ, consider the fake degree sequence

(2) bλ,k ∶= #{T ∈ SYT(λ) ∶ maj(T ) = k} for k = 0,1,2, . . .

The fake degrees have appeared in a variety of algebraic and representation-theoretic contexts including Green’s work on the degree polynomials of uni-potent GLn(Fq)-representations [Gre55, Lemma 7.4], the irreducible decom-position of type A coinvariant algebras [Sta79, Prop. 4.11], Lusztig’s workon the irreducible representations of classical groups [Lus77], and branchingrules between symmetric groups and cyclic subgroups [Ste89, Thm. 3.3]. Theterm “fake degree” was apparently coined by Lusztig [Car89], perhaps because# SYT(λ) = ∑k≥0 bλ,k is the degree of the irreducible Sn-representation indexedby λ, so a q-analog of this number is not itself a degree but related to thedegree.

We consider three natural enumerative questions involving the fake degrees:

(I) which bλ,k are zero?(II) are the fake degree sequences unimodal?

(III) are there efficient asymptotic estimates for bλ,k?

We completely settle (I) with the following result. Denote by λ′ the conjugatepartition of λ, and let b(λ) ∶= ∑i≥1(i − 1)λi.

Theorem 1.1. For every partition λ ⊢ n ≥ 1 and integer k such that b(λ) ≤ k ≤(n2)−b(λ′), we have bλ,k > 0 except in the case when λ is a rectangle with at least

two rows and columns and k is either b(λ) + 1 or (n2) − b(λ′) − 1. Furthermore,bλ,k = 0 for k < b(λ) or k > (n2) − b(λ′).

As a consequence of the proof of Theorem 1.1, we identify two rankedposet structures on SYT(λ) where the rank function is determined by maj.Furthermore, as a corollary of Theorem 1.1 we have a new proof of a completeclassification due to the third author [Swa18, Thm. 1.4] generalizing an earlierresult of Klyachko [Kly74] for when the counts

aλ,r ∶= #{T ∈ SYT(λ) ∶ maj(T ) ≡n r}

for λ ⊢ n are nonzero.The easy answer to question (II) is “no”. The fake degree sequences are

not always unimodal. For example, SYT(4,2)maj(q) is not unimodal. SeeExample 2.13. Nonetheless, certain inversion number generating functions

p(k)α (q) which appear in a generalization of SYT(λ)maj(q) are in fact unimodal;

see Definition 7.7 and Corollary 7.10. Furthermore, computational evidencesuggests SYT(λ)maj(q) is typically not far from unimodal.

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Questions (II) and (III) are addressed in a separate article [BKS20a]. Inparticular, we show in that article that the coefficients of SYT(λ(i))maj(q) areasymptotically normal for any sequence of partitions λ(1), λ(2), . . . such thataft(λ(i)) approaches infinity where aft(λ) is the number of boxes outside thefirst row or column, whichever is smaller. The aft statistics on partitions is inFindStat as [RS+18, St001214].

We note that there are polynomial expressions for the fake degrees bλ,k interms of parameters Hi, the number of cells of λ with hook length equal to i.These polynomials are closely related to polynomials that express the numberof permutations Sn of a given inversion number k ≤ n as a function of n bywork of Knuth. See Lemma 3.1 and Corollary 3.3. These polynomials areuseful in some cases, however, we find that in practice Stanley’s formula is themost effective way to compute a given fake degree sequence for partitions up tosize 200. See Remark 2.12 for more on efficient computation using cyclotomicpolynomials.

Symmetric groups are the finite reflection groups of type A. The classifi-cation and invariant theory of both finite irreducible real reflection groupsand complex reflection groups developed over the past century builds on ourunderstanding of the type A case [Hum90]. In particular, these groups areclassified by Shephard–Todd into an infinite family G(m,d,n) together with 34exceptions. Using work of Stembridge on generalized exponents for irreduciblerepresentations, the analog of (1) can be phrased for all Shephard–Todd groupsas

(3) g{λ}d(q) ∶= #{λ}

d

d⋅ [ nα(λ)

]q;d

⋅m

∏i=1

SYT(λ(i))maj(qm) =∑ b{λ}d,kqk

where λ = (λ(1), . . . , λ(m)) is a sequence of m partitions with n cells total,α(λ) = (∣λ(1)∣, . . . , ∣λ(m)∣) ⊧ n, d ∣m, and {λ}d is the orbit of λ under the groupCd of (m/d)-fold cyclic rotations; see Corollary 8.2. The polynomials [ nα(λ)]q;dare deformations of the usual q-multinomial coefficients which we explore inSection 7. The coefficients b{λ}d,k are the fake degrees in this case.

We use (3) and Theorem 1.1 to completely classify all nonzero fake degrees forcoinvariant algebras for all Shephard–Todd groups G(m,d,n), which includesthe finite real reflection groups in types A, B, and D. See Corollary 6.4 andCorollary 8.4 for the type B and D cases, respectively. See Theorem 6.3 andTheorem 8.3 for the general Cm ≀ Sn and G(m,d,n) cases, respectively.

The rest of the paper is organized as follows. In Section 2, we give back-ground on tableau combinatorics, Shephard–Todd groups, and their irreduciblerepresentations. Section 3 describes the polynomial formulas for fake degreesin type A. Section 4 presents our combinatorial argument proving Theorem 1.1and giving poset structures on tableaux of a given shape. Section 5 uses the


argument in Section 4 to answer in the affirmative a question of Adin–Elizalde–Roichman about internal zeros of SYT(λ)des(q); see Corollary 5.3. In Section 6,we begin to address the question of characterizing nonzero fake degrees bystarting with the wreath products Cm ≀ Sn = G(m,1, n); see Theorem 6.3. InSection 7, we define the deformed q-multinomials [nα]q;d as rational functionsand give a summation formula, Theorem 7.6, which shows they are polynomial.Finally, in Section 8, we complete the classification of nonzero fake degrees forG(m,d,n) and spell out how (3) relates to Stembridge’s original generatingfunction for the fake degrees in G(m,d,n); see Theorem 8.3 and Corollary 8.2.We discuss potential algebraic and geometric directions for future work inSection 9.

2. Background

In this section, we review some standard terminology and results on combi-natorial statistics and tableaux. Many further details in this area can be foundin [Sta12, Sta99]. We also review background on the finite complex reflectiongroups and their irreducible representations. Further details in this area canbe found in [Car89, Sag91].

2.1. Word and Tableau Combinatorics. Here we review standard combi-natorial notions related to words and tableaux.

Definition 2.1. Given a word w = w1w2⋯wn with letters wi ∈ Z≥1, the contentof w is the sequence α = (α1, α2, . . .) where αi is the number of times i appearsin w. Such a sequence α is called a (weak) composition of n, written as α ⊧ n.Trailing 0’s are often omitted when writing compositions, so α = (α1, α2, . . . , αm)for some m. Note, a word of content (1,1, . . . ,1) ⊧ n is a permutation in thesymmetric group Sn written in one-line notation. The inversion number of wis

inv(w) ∶= #{(i, j) ∶ i < j,wi > wj}.The descent set of w is

Des(w) ∶= {0 < i < n ∶ wi > wi+1}and the major index of w is

maj(w) ∶= ∑i∈Des(w)

i.

The study of permutation statistics is a classical topic in enumerative combi-natorics. The major index statistic on permutations was introduced by PercyMacMahon in his seminal works [Mac13, Mac17]. At first glance, this functionon permutations may be unintuitive, but it has inspired hundreds of papers andmany generalizations; for example on Macdonald polynomials [HHL05], posets

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[ER15], quasisymmetric functions [SW10], cyclic sieving [RSW04, AS17], andbijective combinatorics [Foa68, Car75].

Definition 2.2. Given a finite set W and a function stat∶W → Z≥0, write thecorresponding ordinary generating function as

W stat(q) ∶= ∑w∈W

qstat(w).

Definition 2.3. Let α = (α1, . . . , αm) ⊧ n. We use the following standardq-analogues:

[n]q ∶= 1 + q +⋯ + qn−1 = qn−1q−1 , (q-integer)

[n]q! ∶= [n]q[n − 1]q⋯[1]q, (q-factorial)

(nk)q

∶= [n]q ![k]q ![n−k]q !

∈ Z≥0[q], (q-binomial)

(nα)q

∶= [n]q ![α1]q !⋯[αm]q !

∈ Z≥0[q] (q-multinomial).

Example 2.4. The identity statistic on the setW = {0, . . . , n−1} has generatingfunction [n]q. The “sum” statistic on W = ∏nj=1{0, . . . , j − 1} has generatingfunction [n]q!. It is straightforward to show that also Sinvn ∶= ∑w∈Sn qinv(w) =[n]q!.

For α ⊧ n, let Wα denote the set of all words of content α. A classic resultof MacMahon is that maj and inv have the same distribution on Wα which isdetermined by the corresponding q-multinomial.

Theorem 2.5. [Mac17, §1] For each α ⊧ n,

Wmajα (q) = (n

α)q

= Winvα (q).(4)

Definition 2.6. A polynomial P (q) = ∑ni=0 ciqi of degree n is symmetric ifci = cn−i for 0 ≤ i ≤ n. We generally say P (q) is symmetric also if there existsan integer k such that qkP (q) is symmetric. We say P (q) is unimodal if

c0 ≤ c1 ≤ ⋯ ≤ cj ≥ cj+1 ≥ ⋯ ≥ cnfor some 0 ≤ j ≤ n. Furthermore, P (q) has no internal zeros provided thatcj ≠ 0 whenever ci, ck ≠ 0 and i < j < k.

From Theorem 2.5 and the definition of the q-multinomials, we see that eachWmajα (q) is a symmetric polynomial with constant and leading coefficient 1.Indeed, these polynomials are unimodal generalizing the well-known case forGaussian coefficients [Sta80, Thm 3.1] and [Zei89]. It also follows easily fromMacMahon’s theorem that Wmajα (q) has no internal zeros.


2.2. Partitions and Standard Young Tableaux.

Definition 2.7. A composition λ ⊧ n such that λ1 ≥ λ2 ≥ . . . is called apartition of n, written as λ ⊢ n. The size of λ is ∣λ∣ ∶= n and the length `(λ) ofλ is the number of non-zero entries. The Young diagram of λ is the upper-leftjustified arrangement of unit squares called cells where the ith row from thetop has λi cells following the English notation; see Figure 1a. The cells of atableau are indexed by matrix notation when we refer to their row and column.The hook length of a cell c ∈ λ is the number hc of cells in λ in the same row asc to the right of c and in the same column as c and below c, including c itself;see Figure 1b. A corner of λ is any cell with hook length 1. A notch of λ isany (i, j) not in λ such that both (i − 1, j) and (i, j − 1) are in λ. Note thatnotches cannot be in the first row or column of λ. A bijective filling of λ is anylabeling of the cells of λ by the numbers [n] = {1,2, . . . , n}. The symmetricgroup Sn acts on bijective fillings of λ by acting on the labels.

(a) Young diagram of λ.

8 7 6 3 2 14 3 23 2 1

(b) Hook lengths of λ.

Figure 1. Constructions related to the partition λ = (6,3,3) ⊢12. The partition has corners at positions (3,3) and (1,6) andone notch at position (2,4).

Definition 2.8. A skew partition λ/ν is a pair of partitions (ν, λ) such thatthe Young diagram of ν is contained in the Young diagram of λ. The cellsof λ/ν are the cells in the diagram of λ which are not in the diagram of ν,written c ∈ λ/ν. We identify straight partitions λ with skew partitions λ/∅where ∅ = (0,0, . . .) is the empty partition. The size of λ/ν is ∣λ/ν∣ ∶= ∣λ∣ − ∣ν∣.The notions of bijective filling, hook lengths, corners, and notches naturallyextend to skew partitions as well.

Definition 2.9. Given a sequence of partitions λ = (λ(1), . . . , λ(m)), we identifythe sequence with the block diagonal skew partition obtained by translating theYoung diagrams of the λ(i) so that the rows and columns occupied by thesecomponents are disjoint, form a valid skew shape, and they appear in orderfrom top to bottom as depicted in Figure 2.

Definition 2.10. A standard Young tableau of shape λ/ν is a bijective fillingof the cells of λ/ν such that labels increase to the right in rows and down

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Figure 2. Diagram for the skew partition λ/ν = 76443/4433,which is also the block diagonal skew shape λ =((3,2), (1,1), (3)).

columns; see Figure 3. The set of standard Young tableaux of shape λ/ν isdenoted SYT(λ/ν). The descent set of T ∈ SYT(λ/ν) is the set Des(T ) of alllabels i in T such that i + 1 is in a strictly lower row than i. The major indexof T is

maj(T ) ∶= ∑i∈Des(T )

i.

1 2 4 7 9 123 6 105 8 11

2 64 5

1 3 7

Figure 3. On the left is a standard Young tableau of straightshape λ = (6,3,3) with descent set {2,4,7,9,10} and majorindex 32. On the right is a standard Young tableau of blockdiagonal skew shape (7, 5, 3)/(5, 3) corresponding to the sequenceof partitions ((2), (2), (3)) with descent set {2,6} and majorindex 8.

The block diagonal skew partitions λ allow us to simultaneously considerwords and tableaux as follows. Let Wα be the set of all words with contentα = (α1, . . . , αk). Letting λ = ((αk), . . . , (α1)), we have a bijection

(5) φ∶SYT(λ) ∼→Wαwhich sends a tableau T to the word whose ith letter is the row number inwhich i appears in T , counting from the bottom up rather than top down.For example, using the skew tableau T on the right of Figure 3, we haveφ(T ) = 1312231 ∈ W(3,2,2). It is easy to see that Des(φ(T )) = Des(T ), so thatmaj(φ(T )) = maj(T ).

2.3. Major Index Generating Functions. Stanley gave the following an-alogue of Theorem 2.5 for standard Young tableaux of a given shape. It


generalizes the famous Frame–Robinson–Thrall Hook-Length Formula [FRT54,Thm. 1] or [Sta99, Cor. 7.21.6] obtained by setting q = 1.

Theorem 2.11. [Sta99, 7.21.5] Let λ ⊢ n with λ = (λ1, λ2, . . .). Then

(6) SYT(λ)maj(q) =qb(λ)[n]q!∏c∈λ[hc]q

where b(λ) ∶= ∑(i − 1)λi and hc is the hook length of the cell c.

Remark 2.12. Since # SYT(λ) typically grows extremely quickly, Stanley’sformula offers a practical way to compute SYT(λ)maj(q) even when n ≈ 100 byexpressing both the numerator and denominator, up to a q-shift, as a productof cyclotomic polynomials and canceling all factors from the denominator. Weprefer to use cyclotomic factors over linear factors in order to avoid arithmeticin cyclotomic fields.

Example 2.13. For λ = (4,2), b(λ) = 2 and the multiset of hook lengths is{12,22,4,5} so ∣SYT(λ)∣ = 9 by the Hook-Length Formula. The major indexgenerating function is given by

SYT(4,2)maj(q) = q8 + q7 + 2q6 + q5 + 2q4 + q3 + q2

= q2[6]q!

[5]q[4]q[2]q[2]q= q2

[6]q[3]q[2]q

.

Note, SYT(4,2)maj(q) is symmetric but not unimodal.For λ = (4,2,1), b(λ) = 4 and the multiset of hook lengths is {13,2,3,4,6}

so ∣SYT(λ)∣ = 35 by the Hook-Length Formula. The major index generatingfunction is given by

SYT(4,2,1)maj(q) = q14 + 2q13 + 3q12 + 4q11 + 5q10 + 5q9 + 5q8 + 4q7 + 3q6

+ 2q5 + q4 = q4[7]q!

[6]q[4]q[3]q[2]q= q4[7]q[5]q.

Note, SYT(4,2,1)maj(q) is symmetric and unimodal.

Example 2.14. We recover q-integers, q-binomials, and q-Catalan numbers, upto q-shifts as special cases of the major index generating function for tableauxas follows:

SYT(λ)maj(q) =

⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩

q[n]q if λ = (n,1),q(k+12

)(nk)q

if λ = (n − k + 1,1k),qn 1

[n+1]q(2nn)q

if λ = (n,n).

The following strengthening of Stanley’s formula to λ is well known (e.g. see[Ste89, (5.6)]), though since it is somewhat difficult to find explicitly in theliterature, we include a short proof.

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Theorem 2.15. Let λ = (λ(1), . . . , λ(m)) where λ(i) ⊢ αi and n = α1 +⋯ + αm.Then

(7) SYT(λ)maj(q) = ( nα1, . . . , αm

)q

⋅m

∏i=1

SYT(λ(i))maj(q).

Proof. The stable principal specialization of skew Schur functions is given by

sλ/ν(1, q, q2, . . .) =SYT(λ/ν)maj(q)∏∣λ/ν∣j=1 (1 − qj)

;

see [Ste89, Lemma 3.1] or [Sta99, Prop.7.19.11]. On the other hand, it is easyto see from the definition of a skew Schur function as the content generatingfunction for semistandard tableaux of the given shape that

sλ(x1, x2, . . .) =m

∏i=1

sλ(i)(x1, x2, . . .).

The result quickly follows. �

Remark 2.16. Theorem 2.11 and Theorem 2.15 have several immediate corol-laries. First, we recover MacMahon’s result, Theorem 2.5, from Theorem 2.15when λ = ((αm), (αm−1), . . .) by using the maj-preserving bijection φ in (5).Second, each SYT(λ)maj(q) is symmetric (up to a q-shift) with leading coef-ficient 1. In particular, there is a unique “maj-minimizer” and “maj-maximizer”tableau in each SYT(λ). Moreover,(8) min maj(SYT(λ)) = b(λ)and

(9) max maj(SYT(λ)) = (n2) − b(λ′) = b(λ) + (∣λ∣ + 1

2) −∑

c∈λ

hc

where b(λ) ∶= ∑i b(λ(i)) and b(λ′) ∶= ∑i b(λ(i) ′).

For general skew shapes, SYT(λ/ν)maj(q) does not factor as a productof cyclotomic polynomials times q to a power. A “q-Naruse” formula dueto Morales–Pak–Panova, [MPP15, (3.4)], gives an analogue of Theorem 2.11involving a sum over “excited diagrams,” though the resulting sum has a singleterm precisely for the block diagonal skew partitions λ.

2.4. Complex Reflection Groups. A complex reflection group is a finitesubgroup of GL(Cn) generated by pseudo-reflections, which are elements whichpointwise fix a codimension-1 hyperplane. Shephard–Todd, building on workof Coxeter and others, famously classified the complex reflection groups [ST54].The irreducible representations were constructed by Young, Specht, Lusztig,and others. We now summarize these results and fix some notation.


Definition 2.17. A pseudo-permutation matrix is a matrix where each rowand column has a single non-zero entry. For positive integers m,n, the wreathproduct Cm ≀ Sn ⊂ GL(Cn) is the group of n × n pseudo-permutation matriceswhose non-zero entries are complex mth roots of unity. For d ∣m, let G(m,d,n)be the Shephard–Todd group consisting of matrices x ∈ Cm ≀ Sn where theproduct of the non-zero entries in x is an (m/d)th root of unity. In fact,G(m,d,n) is a normal subgroup of Cm ≀ Sn of index d with cyclic quotient(Cm ≀ Sn)/G(m,d,n) ≅ Cd of order d.Theorem 2.18. [ST54] Up to isomorphism, the complex reflection groups areprecisely the direct products of the groups G(m,d,n), along with 34 exceptionalgroups.

Remark 2.19. Special cases of the Shephard–Todd groups include the fol-lowing. The Weyl group of type An−1, or equivalently the symmetric groupSn, is isomorphic to G(1, 1, n). The Weyl groups of both types Bn and Cn areG(2,1, n), the group of n × n signed permutation matrices. The subgroup ofthe group of signed permutations whose elements have evenly many negativesigns is the Weyl group of type Dn, or G(2,2, n) as a Shephard–Todd group.We also have that G(m,m, 2) is the dihedral group of order 2m, and G(m, 1, 1)is the cyclic group Cm of order m.

The complex irreducible representations of Sn were constructed by Young[You77] and are well known to be certain modules Sλ canonically indexed bypartitions λ ⊢ n. These representations are beautifully described in [Sag91].Specht extended the construction to irreducibles for G ≀ Sn where G is a finitegroup.

Theorem 2.20. [Spe35] The complex inequivalent irreducible representationsof Cm ≀ Sn are certain modules Sλ indexed by the sequences of partitions λ =(λ(1), . . . , λ(m)) for which ∣λ∣ ∶= ∣λ(1)∣ +⋯ + ∣λ(m)∣ = n.Remark 2.21. The version we give of Theorem 2.20 was stated by Stembridge[Ste89, Thm. 4.1]. The Cm-irreducibles are naturally though non-canonicallyindexed by Z/m up to one of φ(m) additive automorphisms, where φ(m)is Euler’s totient function. Correspondingly, one may identify Z/m with{1, . . . ,m} and obtain φ(m) different indexing schemes for the Cm ≀ Sn-irreduc-ibles. The resulting indexing schemes are rearrangements of one another, andour results will be independent of these choices.

Clifford described a method for determining the branching rules of irre-ducible representations for a normal subgroup of a given finite group [Cli37].Stembridge combined this method with Specht’s theorem to describe the irreduc-ible representations for all Shephard–Todd groups from the Cm ≀ Sn-irreduciblerepresentations.

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We use Stembridge’s terminology where possible. In particular, for d ∣ m,the (m/d)-fold cyclic rotations are the elements in the subgroup isomorphic toCd of Sm generated by σ

m/dm , where σm = (1, 2, . . . ,m) is the long cycle. Let Sm

act on block diagonal partitions of the form λ = (λ(1), . . . , λ(m)) by permutingthe blocks. This action restricts to Cd = ⟨σm/dm ⟩ as well. Let {λ}d denote theorbit of λ under the (m/d)-fold cyclic rotations in Cd. Note, the number ofblock diagonal partitions in such a Cd-orbit, denoted #{λ}d, always divides d,but could be less than d if λ contains repeated partitions.

For example, take d = 2 and m = 6. If λ = ((1), (2), (3,2), (4), (5), (6,1)),then {λ}2 has two elements, λ and ((4), (5), (6,1), (1), (2), (3,2)). If µ =((1), (2), (3,2), (1), (2), (3,2)), then {µ}2 only contains the element µ.

Theorem 2.22. [Ste89, Remark after Prop. 6.1] The complex inequivalentirreducible representations of G(m,d,n) are certain modules S{λ}d,c indexed bythe pairs ({λ}d, c) where λ = (λ(1), . . . , λ(m)) is a sequence of partitions with∣λ∣ = n, {λ}d is the orbit of λ under (m/d)-fold cyclic rotations, and c is anypositive integer 1 ≤ c ≤ d

#{λ}d.

Remark 2.23. As with Cm ≀ Sn, the indexing scheme is again non-canonicalin general up to a choice of orbit representative, though our results relyingon this work are independent of these choices. In fact, Stembridge usesλ = (λ(m−1), . . . , λ(0)), which is the most natural setting for Theorem 2.22and Theorem 2.39 below. The fake degrees for irreducibles Sλ of Cm ≀ Sn areinvariant up to a q-shift under all permutations of λ in Sm, so for our purposesthe indexing scheme is largely irrelevant. The fake degrees for irreduciblesS{λ}

d,c of G(m,d,n), however, are only invariant under the (m/d)-fold cyclicrotations of λ in general. In this case, strictly speaking our λ(i) correspondsto the irreducible cyclic group representation χi−1 defined by χi−1(σm) = ωi−1mwhere ωm is a fixed primitive mth root of unity in the sense that

Sλ ∶= (χ0 ≀ Sλ(1) ⊗⋯⊗ χm−1 ≀ Sλ(m))↑Cm≀SnCm≀Sα(λ) ;

see [Ste89, (4.1)]. Since we have no need of these explicit representations, wehave used the naive indexing scheme throughout.

Example 2.24. For the type Bn group G(2,1, n), the irreducible representa-tions are indexed by pairs (λ,µ) since C1 is the trivial group and so in eachcase c = 1.

Example 2.25. For the type Dn group G(2,2, n), the irreducible representa-tions can be thought of as being indexed by the sets {λ,µ} with λ ≠ µ and∣λ∣ + ∣µ∣ = n together with the pairs (ν,1) and (ν,2) where ν ⊢ n/2. The orbitsalone can be thought of as the 2 element multisets {λ,µ} with ∣λ∣ + ∣µ∣ = n.


2.5. Coinvariant Algebras. As mentioned in the introduction, Stanley (see[Sta79]) and Lusztig (unpublished) determined the graded irreducible decom-position of the type A coinvariant algebra via the major index generatingfunction on standard Young tableaux. Stembridge was the first to publish acomplete proof of this result and extended it to the complex reflection groupsG(m,d,n) [Ste89]. We now summarize these results.

Definition 2.26. Any group G ⊂ GL(Cn) acts on the polynomial ring with nvariables C[x1, . . . , xn] by identifying Cn with SpanC{x1, . . . , xn} and extendingthe G-action multiplicatively. The coinvariant algebra of G is the quotient ofC[x1, . . . , xn] by the ideal generated by homogeneous G-invariant polynomialsof positive degree, which is thus a graded G-module.

Definition 2.27. Let Rn denote the coinvariant algebra of Sn. For λ ⊢ n, letgλ(q) be the fake degree polynomial whose kth coefficient is the multiplicity ofSλ in the kth degree piece of Rn.

Theorem 2.28 (Lusztig–Stanley, [Sta79, Prop. 4.11]). For a partition λ,

gλ(q) = SYT(λ)maj(q).

Equivalently, the multiplicity of Sλ in the kth degree piece of the type Acoinvariant algebra Rn is bλ,k, the number of standard tableaux of shape λ ⊢ nwith major index k.

Definition 2.29. Let Rm,n denote the coinvariant algebra of Cm ≀ Sn. Set

bλ,k ∶= the multiplicity of Sλ in the kth degree piece of Rm,n.

Write the corresponding fake degree polynomial as

gλ(q) ∶=∑k

bλ,kqk.

Definition 2.30. Given a sequence of partitions λ = (λ(1), . . . , λ(m)), recall

b(α(λ)) =m

∑i=1

(i − 1)∣λ(i)∣.

We continue to identify λ with a block diagonal skew partition when conve-nient, as in Definition 2.9. Thus, SYT(λ) is the set of standard Young tableauxon the block diagonal skew partition λ. We will abuse notation and defineb(α(T )) ∶= b(α(λ)) for any T ∈ SYT(λ), which is not necessary in the nexttheorem but will be essential for the general Shephard–Todd groups G(m,d,n).

Theorem 2.31. [Ste89, Thm. 5.3] For λ = (λ(1), . . . , λ(m)) with ∣λ∣ = n,

gλ(q) = qb(α(λ)) SYT(λ)maj(qm).

13

Equivalently, the multiplicity of Sλ in the kth degree piece of the Cm ≀ Sncoinvariant algebra Rm,n is the number of standard tableaux T of block diagonalshape λ with k = b(α(T )) +m ⋅maj(T ).

Remark 2.32. By (7), we have an explicit product formula for gλ(q) also.Furthermore, in [BKS20a], we characterize the possible limiting distributionsfor the coefficients of the polynomials SYT(λ)maj(q). We show that in mostcases, the limiting distribution is the normal distribution. Consequently,that characterization can be interpreted as a statement about the asymptoticdistribution of irreducible components in different degrees of the Cm ≀ Sncoinvariant algebras.

Corollary 2.33. In type Bn, the irreducible indexed by (λ,µ) with ∣λ∣ = k and∣µ∣ = n − k has fake degree polynomial

g(λ,µ)(q) = q∣µ∣+2b(λ)+2b(µ)(nk)q2

[k]q2 !∏c∈λ[hc]q2

[n − k]q2 !∏c′∈µ[hc′]q2

.

Definition 2.34. Let Rm,d,n denote the coinvariant algebra of G(m,d,n)assuming d ∣m. For an orbit {λ}d of a sequence of m partitions with n totalcells under (m/d)-fold cyclic rotations, set

b{λ}d,k ∶= the multiplicity of S{λ}d,c in the kth degree piece of Rm,d,n,

which in fact depends only on the orbit {λ}d and not the number c by [Ste89,Prop. 6.3]. Write the corresponding fake degree polynomial as

g{λ}d(q) ∶=∑

k

b{λ}d,kqk.

Theorem 2.35. [Ste89, Cor. 6.4] Let {λ}d be the orbit of a sequence of mpartitions λ with ∣λ∣ = n under (m/d)-fold cyclic rotations. Then

g{λ}d(q) = ({λ}

d)b○α (q)[d]qnm/d

SYT(λ)maj(qm)

where

({λ}d)b○α (q) ∶= ∑µ∈{λ}d

qb(α(µ)).

Corollary 2.36 ([Lus77, Sect. 2.5], [Ste89, Cor. 6.5]). In type Dn, anirreducible indexed by {λ}2 with λ = (λ,µ) and ∣λ∣ = k, ∣µ∣ = n − k has fakedegree polynomial

g{λ}2(q) = κλµq2b(λ)+2b(µ)

qk + qn−k1 + qn

(nk)q2

[k]q2 !∏c∈λ[hc]q2

[n − k]q2 !∏c′∈µ[hc′]q2

,

where κλµ = 1 if λ ≠ µ and κλλ = 1/2.


Observe that Theorem 2.31 gives a direct tableau interpretation of thecoefficients of gλ(q). More generally, Stembridge gave a tableau interpretationof the coefficients of g{λ}

d(q) which we next describe.

Definition 2.37. For a given m,d,n, let λ = (λ(1), . . . , λ(m)) be a sequence of mpartitions with ∣λ∣ = n. Let {λ}d be the orbit of λ under (m/d)-fold rotations.The cyclic group Cd = ⟨σm/dm ⟩ acts on the disjoint union ⊔µ∈{λ}d SYT(µ) asfollows. Given µ = (µ(1), . . . , µ(m)) ∈ {λ}d, each T ∈ SYT(µ) may be consideredas a sequence T = (T (1), . . . , T (m)) of fillings of the shapes µ(i). The group Cdacts by (m/d)-fold rotations of this sequence of fillings. Write the resultingorbit as {T}d, which necessarily has size d. For such a T , the largest entry ofT , namely n, appears in some T (k). If among the elements of the orbit {T}d ofT this value k is minimal for T itself, then we call T the canonical standardtableau representative for {T}d. Let

SYT({λ}d) ⊂ ⊔µ∈{λ}d

SYT(µ)

be the set of canonical standard tableau representatives of orbits {T}d forT ∈ SYT(λ). Recall, b(α(T )) ∶= b(α(µ)) = ∑(i− 1)∣µ(i)∣ if T ∈ SYT(µ), so b ○αis not generally constant on SYT({λ}d).

Remark 2.38. When the parts λ(i) are all non-empty, the set SYT({λ}d) isthe set of standard block diagonal skew tableaux of some shape µ ∈ {λ}d wheren = ∣λ∣ is in the upper-right-most partition possible among the (m/d)-fold cyclicrotations of its blocks. Since every orbit {T}d has size d, we have

# SYT({λ}d) = #{λ}d

d# SYT(λ).

Theorem 2.39. [Ste89, Thm. 6.6] Let λ be a sequence of m partitions with∣λ∣ = n. Let {λ}d be the orbit of λ under (m/d)-fold cyclic rotations. Then

g{λ}d(q) = SYT({λ}d)b○α+m⋅maj(q).

Equivalently, the multiplicity of S{λ}d,c in the kth degree piece of the G(m,d,n)

coinvariant algebra Rm,d,n is the number of canonical standard tableaux T ∈SYT({λ}d) with k = b(α(T )) +m ⋅maj(T ).

3. Polynomial Formulas For Fake Degrees

In this section, we briefly show how to construct polynomial formulas forthe fake degrees bλ,k directly from Stanley’s q-hook length formula. We willuse these polynomials in the next section for small changes from the minimalmajor index. Our results extend to a formula for counting permutations of agiven inversion number.

15

Given λ, let

Hi(λ) = #{c ∈ λ ∶ hc = i},(10)mi(λ) = #{k ∶ λk = i}.(11)

If λ is understood, we abbreviate Hi =Hi(λ). For any nonnegative integer kand polynomial f(q), let [qk]f(q) be the coefficient of qk in f(q).

Lemma 3.1. For every λ ⊢ n and k = b(λ) + d, we have

(12) bλ,k = [qb(λ)+d]SYT(λ)maj(q) = ∑µ⊢dµ1≤n

∣λ∣

∏i=1

(Hi +mi(µ) − 2mi(µ)

)

which is a polynomial in the Hi’s for every positive integer n.

Proof. By Theorem 2.11, we have

(13) q−b(λ) SYT(λ)maj(q) =[n]q!∏c∈λ[hc]q

=n

∏i=1

(1 − qi)−(Hi−1).

The result follows using the expansion (1 − qi)−j = ∑∞n=0 (j+n−1n

)qin and multipli-cation of ordinary generating functions. �

Note that if Hi(λ) = 0 and mi(µ) = 1, then the corresponding binomial coef-ficient in (12) is −1, so it is not obvious from this formula that the coefficientsbλ,k are all nonnegative, which is clearly true by definition.

Remark 3.2. The first few polynomials are given by

[qb(λ)+1]SYT(λ)maj(q) =H1 − 1= #{notches of λ},

[qb(λ)+2]SYT(λ)maj(q) = (H12

) +H2 − 1,

[qb(λ)+3]SYT(λ)maj(q) = (H1 + 13

) + (H1 − 1)(H2 − 1) + (H3 − 1)

[qb(λ)+4]SYT(λ)maj(q) = (H1 + 24

) + (H22

) + (H12

)(H2 − 1)

+ (H1 − 1)(H3 − 1) + (H4 − 1).These exact formulas hold for all ∣λ∣ ≥ 4. For smaller size partitions some termswill not appear.

It is interesting to compare these polynomials to the ones described by Knuthfor the number of permutations with k ≤ n inversions in Sn in [Knu73, p.16].See also [Sta12, Ex. 1.124] and [OEI18, A008302]. We can extend Knuth’sformulas to all 0 ≤ k ≤ (n2) using the same idea.


Corollary 3.3. For fixed positive integers k and n, we have

(14) #{w ∈ Sn ∶ inv(w) = d} =∑(−1)#{µi>1}(n +m1(µ) − 2

m1(µ))

where the sum is over all partitions µ ⊢ d such that µ1 ≤ n and all of the partsof µ larger than 1 are distinct.

The proof follows in exactly the same way from the formula

∑w∈Sn

qinv(w) =n

∏i=1

[i]q =n

∏i=1

[i]q/[1]q = (1 − q)−nn

∏i=1

(1 − qi).

In essence, this is the case of the q-hook length formula when all of the hooksare of length 1.

Remark 3.4. Let T (d,n) be the number of partitions µ ⊢ d such that µ1 ≤ nand all of the parts of µ larger than 1 are distinct. The triangle of numbersT (d,n) for 1 ≤ n ≤ d is [OEI18, A318806].

4. Type A Internal Zeros Classification

As a corollary of Stanley’s formula, we know that for every partition λ ⊢ n ≥ 1there is a unique tableau with minimal major index b(λ) and a unique tableauwith maximal major index (n2)− b(λ′). These two agree for shapes consisting ofone row or one column, and otherwise they are distinct. It is easy to identifythese two tableaux in SYT(λ); see Definition 4.1 below. Then, we classify allof the values k such that b(λ) < k < (n2)− b(λ′) and the fake degree bλ,k = 0. Werefer to such k as internal zeros, meaning the location of zeros in the fake degreesequence for λ between the known minimal and maximal nonzero locations.

Definition 4.1.

(1) The max-maj tableau for λ is obtained by filling the outermost, max-imum length, vertical strip in λ with the largest possible numbers∣λ∣, ∣λ∣ − 1, . . . , ∣λ∣ − `(λ) + 1 starting from the bottom row and going up,then filling the rightmost maximum length vertical strip containing cellsnot previously used with the largest remaining numbers, etc.

(2) The min-maj tableau of λ is obtained similarly by filling the outermost,maximum length, horizontal strip in λ with the largest possible num-bers ∣λ∣, ∣λ∣ − 1, . . . , ∣λ∣ − λ1 + 1 going right to left, then filling the lowestmaximum length horizontal strip containing cells not previously usedwith the largest remaining numbers, etc.

See Figure 4 for an example. Note that the max-maj tableau of λ is thetranspose of the min-maj tableau of λ′.

The qb(λ)+1 coefficients of SYT(λ)maj(q) can be computed as in Lemma 3.1or Remark 3.2, resulting in the following.

17

1 2 3 5 9 13

4 6 10 14

7 11 15

8 12 16

17

(a) A max-maj tableau and itsoutermost vertical strip.

1 3 4 11 16 17

2 6 7 15

5 9 10

8 13 14

12

(b) A min-maj tableau and itsoutermost horizontal strip.

Figure 4. Max-maj tableau and min-maj tableau for λ =(6,4,3,3,1).

Corollary 4.2. We have [qb(λ)+1]SYT(λ)maj(q) = 0 if and only if λ is arectangle. If λ is a rectangle with more than one row and column, then[qb(λ)+2]SYT(λ)maj(q) = 1.

A similar statement holds for maj(T ) = (n2) − b(λ′) − 1 by symmetry. Thus,SYTmaj(q) has internal zeros when λ is a rectangle with at least two rows andcolumns. We will show these are the only internal zeros of type A fake degrees,proving Theorem 1.1.

Definition 4.3. Let E(λ) denote the set of exceptional tableaux of shape λconsisting of the following elements.

(i) For all λ, the max-maj tableau for λ.(ii) If λ is a rectangle, the min-maj tableau for λ.(iii) If λ is a rectangle with at least two rows and columns, the unique tableau

in SYT(λ) with major index equal to (n2) − b(λ′) − 2. It is obtained fromthe max-maj tableau of λ by applying the cycle (2, 3, . . . , `(λ)+ 1), whichreduces the major index by 2.

For example, E(64331) consists of just the max-maj tableau for 64331 inFigure 4a, while E(555) has the following three elements:

1 2 3 4 56 7 8 9 1011 12 13 14 15

1 2 7 10 133 5 8 11 144 6 9 12 15

1 4 7 10 132 5 8 11 143 6 9 12 15

.

We prove Theorem 1.1 by constructing a map

(15) ϕ ∶SYT(λ) ∖ E(λ)Ð→ SYT(λ)with the property

(16) maj(ϕ(T )) = maj(T ) + 1.For most tableaux T , we can find another tableau T ′ of the same shape suchthat maj(T ′) = maj(T ) + 1 by applying some simple cycle to the values in T ,


meaning a permutation whose cycle notation is either (i, i + 1, . . . , k − 1, k) or(k, k − 1, . . . , i + 1, i) for some i < k. We will show there are 5 additional rulesthat must be added to complete the definition.

We note that technically the symmetric group Sn does not act on SYT(λ)for λ ⊢ n since this action will not generally preserve the row and columnstrict requirements for standard tableaux. However, Sn acts on the set of allbijective fillings of λ using the alphabet {1,2, . . . , n} by acting on the values.We will only apply permutations to tableaux after locating all values in someinterval [i, j] = {i, i+1, . . . , j} in T . The reader is encouraged to verify that thespecified permutations always maintain the row and column strict properties.

4.1. Rotation Rules. We next describe certain configurations in a tableauwhich imply that a simple cycle will increase maj by 1. Recall, the cells of atableau are indexed by matrix notation.

Definition 4.4. Given λ ⊢ n and T ∈ SYT(λ), a positive rotation for T is aninterval [i, k] ⊂ [n] such that if T ′ ∶= (i, i+ 1, . . . , k − 1, k) ⋅T , then T ′ ∈ SYT(λ)and there is some j for which

{j} = Des(T ′) −Des(T ) and {j − 1} = Des(T ) −Des(T ′).

Intuitively, a positive rotation is one for which j − 1 ∈ Des(T ) becomes j ∈Des(T ′) and all other entries remain the same. Consequently, maj(T ′) =maj(T ) + 1. We call j the moving descent for the positive rotation.

The positive rotations can be characterized explicitly as follows. The proofis omitted since it follows directly from the pictures in Figure 5.

Lemma 4.5. An interval [i, k] is a positive rotation for T ∈ SYT(λ) if andonly if i < k and there is some necessarily unique moving descent j with1 ≤ i ≤ j ≤ k ≤ n such that(a) i, . . . , j − 1 form a horizontal strip, j − 1, j form a vertical strip, and j, j +

1, . . . , k form a horizontal strip;(b) if i < j, then i appears strictly northeast of k and i−1 is not in the rectangle

bounding i and k;(c) if i = j, then i − 1 appears in the rectangle bounding i and k;(d) if j < k, then k appears strictly northeast of k − 1 and k + 1 is not in the

rectangle bounding k and k − 1; and(e) if j = k, then k + 1 appears in the rectangle bounding k and k − 1.

See Figure 5 for diagrams summarizing these conditions.

In addition to the positive rotations above, we can also apply negativerotations, which are defined exactly as in Definition 4.4 with (i, i+1, . . . , k−1, k)replaced by (k, k − 1, . . . , i + 1, i) and the rest unchanged. Combinatorially,

19

��XXXi − 1 i ⋯ j − 1k

j ⋯ k − 1 ��XXXk + 1Ð→

��XXXi − 1 i + 1 ⋯ ji

j + 1 ⋯ k ��XXXk + 1

(a) Schematic of a positive rotation with i < j < k.��XXXi − 1 i i + 1 ⋯ k − 1k k + 1 Ð→

��XXXi − 1 i + 1 ⋯ k − 1 ki k + 1

(b) Schematic of a positive rotation with i < j = k.i − 1 ki i + 1 ⋯ k − 1 ��XXXk + 1 Ð→

i − 1 ii + 1 i + 2 ⋯ k ��XXXk + 1

(c) Schematic of a positive rotation with i = j < k.

Figure 5. Summary diagrams for positive rotations.

negative rotations can be obtained from positive rotations by applying inverse-transpose moves, that is, by applying negative cycles (k, k − 1, . . . , i) to thetranspose of the configurations in Figure 5 and reversing the arrows. Explicitly,we have the following analogue of Lemma 4.5. See Figure 6 for the correspondingdiagrams.

Lemma 4.6. An interval [i, k] is a negative rotation for T ∈ SYT(λ) if andonly if i < k and there is some necessarily unique moving descent j with1 ≤ i ≤ j ≤ k ≤ n such that(a) i, . . . , j form a vertical strip, j, j +1 form a horizontal strip, and j +1, . . . , k

form a vertical strip;(b) if i < j, then i + 1 appears strictly southwest of i and i − 1 is not in the

rectangle bounding i and i + 1;(c) if i = j, then i − 1 appears in the rectangle bounding i and i + 1;(d) if j < k, then i appears strictly southwest of k and k + 1 is not in the

rectangle bounding i and k; and(e) if j = k, then k + 1 appears in the rectangle bounding i and k.

Example 4.7. The tableau

1 2 6 7 93 4 8 135 11 12 1510 14


j + 1j + 2⋮k

��XXXi − 1 i ��XXXk + 1i + 1i + 2⋮j

Ð→

jj + 1⋮

k − 1��XXXi − 1 k ��XXXk + 1i

i + 1⋮

j − 1

(a) i < j < k.

��XXXi − 1 ii + 1⋮

k − 1k k + 1

Ð→

��XXXi − 1 ki

i + 1⋮

k − 1 k + 1

(b) i < j = k.

i − 1 i + 1i + 2⋮k

i ��XXXk + 1

Ð→

i − 1 ii + 1⋮

k − 1k ��XXXk + 1

(c) i = j < k.

Figure 6. Summary diagrams for negative rotations.

allows positive rotation rules with [i, k] ∈ {[5,6], [8,9], [8,10], [8,11], [9,13]},and the tableau

1 3 8 10 152 4 9 115 7 13 146 12

allows negative rotation rules with [i, k] ∈ {[4,6], [6,7], [11,12]}.

It turns out that for the vast majority of tableaux, some negative rotationrule applies. The positive rotations can be applied in many of the remainingcases. For example, among the 81,081 tableaux in SYT(5442), there are only24 (i.e., 0.03%) on which we cannot apply any positive or negative rotation rule.For example, no rotation rules can be applied to the following two tableaux:

1 2 3 4 56 7 8 910 11 12 1314 15

and

1 2 3 8 124 6 9 135 7 10 1411 15

.

The following lemma and its corollary give a partial explanation for whynegative rotation rules are so common. Given a tableaux T , let T ∣[z] denotethe restriction of T to those values in [z].

21

Lemma 4.8. Let T ∈ SYT(λ) ∖ E(λ). Suppose z is the largest value such thatT ∣[z] is contained in maxmaj(µ) for some µ. If T ∣[z+1] is not of the form

1 2 ⋯ ii + 1 z + 1i + 2⋮z

then some negative rotation rule applies to T .

Proof. Since T /∈ E(λ), T is not maxmaj(λ), so λ is not a one row or columnshape. We have z ≥ 2 since both two-cell tableaux are the max-maj tableauof their shape. Since maxmaj(µ) is built from successive, outermost, maximallength, vertical strips as in Figure 4a, the same is true of T ∣[z].

First, suppose z is not in the lowest row of T ∣[z]. Let i be the value inthe topmost corner cell in T ∣[z] which is strictly below z. Let j ≥ i be thebottommost cell in the vertical strip of T ∣[z] which contains i. See Figure 7a.We verify the conditions of Lemma 4.6, so the negative [i, z]-rotation ruleapplies with moving descent j. By construction, i, . . . , j form a vertical strip,j, j + 1 form a horizontal strip, and j + 1, . . . , z form a vertical strip. If i < j,then since i is a corner cell, i + 1 appears strictly southwest of i, and i − 1 isabove both i and i + 1 so i − 1 is not in the rectangle bounding i and i + 1. Ifi = j, we see that i − 1 appears in the rectangle bounded by i and i + 1. Wealso see that i appears strictly southwest of z, and z + 1 is not in the rectanglebounding i and z since i is a topmost corner and z is maximal.

Now suppose z is in the lowest row of T ∣[z]. In this case, T ∣[z] is the max-majtableau of its shape, so that z < ∣λ∣ and z + 1 exists in T since T /∈ E(λ). Bymaximality of z, z + 1 cannot be in row 1 or below z. Let i < z be the value inthe rightmost cell of T ∣[z] in the row immediately above z + 1. See Figure 7b.We check that the negative [i, z]-rotation rule applies with moving descentj = z using the conditions in Lemma 4.6. By construction, i, . . . , z form avertical strip. Since z + 1 is not below z, we see that z, z + 1 form a horizontalstrip. Since z + 1 is in the row below i, i+ 1 appears strictly southwest of i. Wealso see that z + 1 appears in the rectangle bounded by i and z by choice ofi. It remains to show that i − 1 is not in the rectangle bounding i and i + 1.Suppose to the contrary that i − 1 is in the rectangle bounding i and i + 1.Then i would have to be in row 1 by choice of i < z. Consequently i + 1 is inrow 2 and strictly west of i, forcing i − 1 to be in row 1 also. It follows fromthe choice of z that T ∣[i] is a single row, the values i, i + 1, . . . z form a verticalstrip, and T ∣[z+1] is of the above forbidden form, giving a contradiction. �Corollary 4.9. If T ∈ SYT(λ) ∖ E(λ) and 1 ∈ Des(T ), then some negativerotation rule applies to T .


1 3 6 112 4 7 125 89 1310

Ð→

1 3 6 102 4 7 115 128 139

(a) For the tableau on the left above, i = 8 and z = 12 since T ∣[12] is contained themax-maj tableau of shape 44322, 12 is not in the lowest row, 8 is in the closet cornerto 12 in T ∣[12] and below 12. Apply the negative rotation (12,11,10,9,8) to get thetableau on the right, and observe maj has increased by 1. The moving descent isj = 10.

1 3 62 4 75 8 11910

Ð→

1 3 62 4 105 7 1189

(b) For the tableau on the left above, i = 7 and z = 10 since T ∣[10] is the max-majtableau of shape 33211, 10 is in the lowest row, 11 is in row 3, and 7 is the largestvalue in T ∣[10] in row 2. Apply the negative rotation (10,9,8,7) to get the tableauon the right, and observe maj has increased by 1. The moving descent is j = z = 10.

Figure 7. Examples of the negative rotations obtained fromLemma 4.8.

Proof. Let z be as in Lemma 4.8. Clearly z ≥ 2 and T ∣[2] is a single column, soT ∣[z+1] cannot possibly be of the forbidden form. �

We also have the following variation on Lemma 4.8. It is based on findingthe largest value q such that T ∣[q] is contained in an exceptional tableau oftype (iii). The proof is again a straightforward verification of the conditions inLemma 4.6, and is omitted.

Lemma 4.10. Let T ∈ SYT(λ) ∖ E(λ). Suppose the initial values of T are ofthe form

1 23 p + 14 ⋮⋮ q⋮ ��HHHq + 1p

or

1 2 ` + 1 ⋯ ⋮ p + 13 z + 1 ⋮ ⋮ ⋮ ⋮4 z + 2 ⋮ ⋮ ⋮ q⋮ ⋮ ⋮ ⋮ ⋮ ��HHHq + 1z ` m ⋯ p

.

In either case, the [p, q]-negative rotation rule applies to T .

23

4.2. Initial Block Rules. Here we describe a collection of five additional blockrules which may apply to a tableau that is not in the exceptional set. In eachcase, if the rule applies, then we specify a permutation of the entries so thatwe either add 1 into the descent set and leave the other descents unchanged, orwe add 1 into the descent set, increase one existing descent by 1, and decreaseone existing descent by 1. Thus, maj will increase by 1 in all cases. Whilethese additional rules are certainly not uniquely determined by these criteria,they are also not arbitrary.

Example 4.11. For a given T ∈ SYT(λ), one may consider all T ′ ∈ SYT(λ)where maj(T ′) = maj(T ) + 1. If T ′ = σ ⋅ T where σ is a simple cycle, then oneof the rotation rules may apply to T . Table 1 summarizes five particular Tfor which no rotation rules apply. These examples have guided our choices indefining the block rules. In all but one of these examples, there is a uniqueT ′ with maj(T ′) = maj(T ) + 1, though in the third case there are two such T ′,one of which ends up being easier to generalize.

Tableau T Tableaux T ′ σ Block rule

1 2 3 74 5 6 8

1 3 4 62 5 7 8 (2,3,4)(6,7) B1

1 2 3 45 6 7

1 3 4 72 5 6 (2,3,4,7,6,5) B2

1 2 34 65 7

1 3 62 45 7

,1 4 52 63 7

(2,3,6,4), (2,4)(3,5) B3, —1 2 73 5 84 6 910

1 4 82 5 93 6 107

(2,4,3)(7,8,9,10) B4

1 23 54 67

1 52 63 74

(2,5,6,7,4,3) B5

Table 1. Some tableaux T ∈ SYT(λ) together with all T ′ =σ ⋅ T ∈ SYT(λ) where maj(T ′) = maj(T ) + 1. See Definition 4.13for an explanation of the final column.

In the remainder of this subsection, we describe the block rules, abbreviatedB-rules. Then, we prove that if no rotation rules are possible for a tableauthen either it is in the exceptional set or we can apply one of the B-rules. TheB-rules cover disjoint cases so no tableau admits more than one block rule. Tostate the B-rules precisely, assume T ∈ SYT(λ) ∖ E(λ) and no rotation ruleapplies.

Notation 4.12. Let c be the largest possible value such that T ∣[c] is containedin the min-maj tableau of a rectangle shape with a columns and b rows.


Consequently, the first a numbers in row i, 1 ≤ i ≤ b−1, of T are (i−1)a+1, . . . , ia,and row b begins with (b − 1)a + 1, (b − 1)a + 2, . . . , c.

Assuming 1 /∈ Des(T ) and T /∈ E(λ), we know a, b ≥ 2 and c ≥ 3. If c + 1 isin T , then it must be either in position (1, a + 1) or (b + 1,1). If c = ab, thenc < ∣λ∣ since T /∈ E(λ), otherwise c = ∣λ∣ is possible. For example, the tableaux

1 2 3 4 5 166 7 8 9 10 1711 12 13 14 15

,1 2 3 4 56 7 8 9 1011 12 13

,1 2 3 45 96 1078

,1 2 7 103 5 8 114 6 9 1213

have (a, b, c) equal to (5,3,15), (5,3,13), (4,2,5), and (2,2,3), respectively.

Definition 4.13. Using Notation 4.12, we identify the block rules with furtherrequired assumptions as follows. See Figure 8 for summary diagrams.

● Rule B1: Assume c = ab, T(1,a+1) = c + 1, T(2,a+1) = c + 2, and a < c − 2.In this case, we perform the rotations (2, . . . , a + 1) and (c, c + 1) whichare sufficiently separated by hypothesis. Then, 1, a + 1 and c becomedescents, and a and c + 1 are no longer descents, so the major index isincreased by 1. The B1 rule is illustrated here with a = 5, b = 3:

B1:1 2 3 4 5 166 7 8 9 10 1711 12 13 14 15

⤿⤿

1 3 4 5 6 152 7 8 9 10 1711 12 13 14 16

The boxed numbers represent descents of the tableau on the left/rightthat are not descents of the tableau on the right/left. The elements notshown (i.e. 18,19, . . . , ∣λ∣) can be in any position.

● Rule B2: Assume c < ab and there exists a 1 ≤ k < a such thatT(b,k) = c and T(b,k+1) ≠ c + 1. In this case, we perform the rotation(2,3, . . . , a,2a,3a, . . . , a(b − 1), c, c − 1, . . . , c − k + 1 = a(b − 1) + 1, a(b −2)+ 1, . . . , 2a+ 1, a+ 1) around the perimeter of T ∣[c]. Now 1 becomes adescent, and the other descents stay the same so the major index againincreases by 1. The B2 rule is illustrated by the following (here a = 5,b = 2 and k = 3):

B2:1 2 3 4 56 7 8 9 1011 12 13 ��ZZ14

⤿1 3 4 5 102 7 8 9 136 11 12 ��ZZ14

The crossed out number 14 means that 14 is not in position (3,4): itcan either be in positions (1,6) or (4,1), or it can be that λ = 553.Again, the numbers 15, . . . , ∣λ∣ can be anywhere in T .

25

● Rule B3: Assume a ≥ 3, c = a + 1, and there exists k ≥ 2 suchthat T(2,2) = a + k + 1, T(3,2) = a + k + 2, and for all i ∈ {1,2, . . . , k}we have T(i+1,1) = a + i. Thus b = 2. Then we apply the rotation(2,3, . . . , a, a + k + 1, a + 1). Now 1 becomes a descent, and the rest ofthe descent set is unchanged so the major index again increases by 1.The B3 rule is illustrated by the following (here a = 4, k = 4):

B3:

1 2 3 45 96 1078

⤿1 3 4 92 56 1078

● Rule B4: Assume that a = 2, c = 3, and there exists k ≥ 2 such that{3,4, . . . , k + 1} appear in column 1 of T , {k + 2, k + 3, . . . ,2k} appearin column 2 in T . Further assume that the set {2k + 1,2k + 2, . . . ,3k}appears in column 3, {3k + 1,3k + 2, . . . ,4k} appears in column 4,etc., until column l for some l > 2 and T(k+1,1) = kl + 1 and T(k+1,2) ≠kl + 2. Thus, b = 2. In this case, we can perform the two rotations(k + 1, k, . . . ,3,2) and (k(l − 1) + 1, k(l − 1) + 2, . . . , kl, kl + 1). Now 1,k + 1 and k(l − 1) enter the descent set, and k and k(l − 1) + 1 leave it,so the major index increases by 1. The B4 rule is illustrated by thefollowing (here k = 3 and l = 4):

B4:

1 2 7 10

3 5 8 114 6 9 1213 ��ZZ14

⤿⤿

1 4 7 112 5 8 12

3 6 9 1310 ��ZZ14

● Rule B5: Assume that a = 2, c = 3, and there exists k > 3 such that{3,4, . . . , k} appear in column 1 of T , {k + 1, k + 2, . . . ,2k − 2} appearin column 2 in T . Furthermore, assume T(k,1) = 2k − 1 and T(k,2) ≠ 2k.Thus, b = 2. Then apply the cycle (k, k−1, . . . , 3, 2, k+1, k+2, . . . , 2k−1)to T . Now 1 becomes a descent, and the rest of the descent set remainsunchanged, so the major index increases by 1. The B5 rule is illustratedby the following (here k = 5):

B5:

1 23 64 75 89 ��ZZ10

⤿1 62 73 84 95 ��ZZ10


Lemma 4.14. If T ∈ SYT(λ), T /∈ E(λ), and 1,2 /∈ Des(T ), then either somerotation rule applies to T or a B1, B2 or B3 rule applies.

Proof. Let c be the largest possible value such that T ∣[c] is contained in the min-maj tableau of a rectangle shape with a columns and b rows, see Notation 4.12.

1 2 ⋯ a ab + 1a + 1 a + 2 ⋯ 2a ab + 2⋮ ⋮ ⋱ ⋮

a(b − 1) + 1 a(b − 1) + 2 ⋯ ab↓

1 3 ⋯ a + 1 ab2 a + 2 ⋯ 2a ab + 2⋮ ⋮ ⋱ ⋮

a(b − 1) + 1 a(b − 1) + 2 ⋯ ab + 1

(a) B1.

1 2 ⋯ ⋯ ⋯ ⋯ aa + 1 a + 2 ⋯ ⋯ ⋯ ⋯ 2a⋮ ⋮ ⋱ ⋮ ⋮ ⋱ ⋮

a(b − 2) + 1 a(b − 2) + 2 ⋯ ⋯ ⋯ ⋯ a(b − 1)a(b − 1) + a a(b − 1) + 2 ⋯ c − 1 c ⋯ ��ZZab

↓

1 3 ⋯ ⋯ ⋯ ⋯ 2a2 a + 2 ⋯ ⋯ ⋯ ⋯ 3a⋮ ⋮ ⋱ ⋮ ⋮ ⋱ ⋮

a(b − 3) + 1 a(b − 2) + 2 ⋯ ⋯ ⋯ ⋯ ca(b − 2) + 1 a(b − 1) + 1 ⋯ c − 2 c − 1 ⋯ ��ZZab

(b) B2.

1 2 ⋯ a − 1 aa + 1 a + k + 1a + 2 a + k + 2⋮

a + k

Ð→

1 3 ⋯ a a + k + 12 a + 1

a + 2⋮

a + k

(c) B3.

27

1 2 2k + 1 ⋯ k(` − 1) + 13 k + 2 2k + 2 ⋯ k(` − 1) + 24 k + 3 2k + 3 ⋯ k(` − 1) + 3⋮ ⋮ ⋮ ⋱ ⋮

k + 1 2k 3k ⋯ k`k` + 1 ��XXXXk` + 2

↓

1 k + 1 2k + 1 ⋯ k(` − 1) + 22 k + 2 2k + 2 ⋯ k(` − 1) + 33 k + 3 2k + 3 ⋯ k(` − 1) + 4⋮ ⋮ ⋮ ⋱ ⋮k 2k 3k ⋯ k` + 1

k(` − 1) + 1 ��XXXXk` + 2

(d) B4.

1 23 k + 14 k + 2⋮ ⋮

k − 1 2k − 3k 2k − 2

2k − 1 ��ZZ2k

Ð→

1 k + 12 k + 23 k + 3⋮ ⋮

k − 2 2k − 2k − 1 2k − 1k ��ZZ2k

(e) B5.

Figure 8. Summary diagrams for block rules.

Since 1,2 /∈ Des(T ) and T /∈ E(λ), we know 1,2,3 are in the first row of T soa ≥ 3, b ≥ 2, and a + 2 ≤ ∣λ∣. By construction, we have T(2,1) = a + 1 and a + 2must appear in position (1, a + 1), (2,2), or (3,1) in T .

Case 1: T(1,a+1) = a + 2. Observe that

T ∣[a+2] =1 2 3 ⋯ a a + 2

a + 1

and z ≥ a + 2. Consequently, T ∣[z+1] cannot be of the form forbidden byLemma 4.8, so a negative rotation rule applies.

Case 2: T(2,2) = a + 2. First suppose c = ab, then T(1,a+1) = c + 1 by choice ofc. Now consider the two subcases, T(2,a+1) = c + 2 and T(2,a+1) ≠ c + 2. In the


former case, as in Figure 8a, the B1 rule applies to T . In the latter case, onemay check that an [i, c + 1]-positive rotation rule applies to T where i = T(b,1).On the other hand, if c < ab, then a B2 rule applies to T as in Figure 8b.Case 3: T(3,1) = a + 2. Let k = min{j ≥ 2 ∣ a + j /∈ Des(T )} so T(k+1,1) = a + kand T(k+2,1) ≠ a + k + 1. Since T /∈ E(λ), we know a + k + 1 exists in T either inposition (1, a + 1) or (2,2), so T ∣[a+k+1] looks like

1 2 3 ⋯ a a + k + 1a + 1a + 2⋮

a + k

or

1 2 3 ⋯ aa + 1 a + k + 1a + 2⋮

a + k

.

If T(1,a+1) = a + k + 1, then Lemma 4.8 shows that a negative rotation ruleapplies to T . On the other hand, if T(2,2) = a + k + 1, then observe that either aB3 move applies or the rotation (a + k, a + k + 1) applies to T , depending onwhether T(3,2) = a + k + 2 or not. �

Lemma 4.15. If T ∈ SYT(λ), T /∈ E(λ), 1 /∈ Des(T ), and 2 ∈ Des(T ), theneither some rotation rule applies to T or a B1, B2, B4 or B5 rule applies.

Proof. Let k = min{j ≥ 3 ∣ j /∈ Des(T )} so the consecutive sequence [3, k]appears in the first column of T and k + 1 does not. By definition of k and thefact that T /∈ E(λ), T must have k+1 in position (1, 3) or (2, 2). If T(1,3) = k+1,then a negative rotation rule holds by Lemma 4.8.

Assume T(2,2) = k+1. Let ` be the maximum value such that [k+1, `] appearsas a consecutive sequence in column 2 of T . If ` < 2(k − 1), then the negativerotation rule for (`, ` − 1, . . . , k) applies to T by the first case of Lemma 4.10.

If ` = 2(k−1) and T(1,3) = `+1, let m be the maximum value such that [`+1,m]appears as a consecutive sequence in column 3 of T . We subdivide on casesfor m again. If m < 3(k − 1), then the negative rotation rule (m,m − 1, . . . , `)applies to T by the second case of Lemma 4.10. If m = 3(k − 1), we considerthe maximal sequence of columns containing a consecutive sequence in rows[1, k − 1] to the right of column 2 until one of two conditions hold

1 2 ` + 1 ⋯ ⋮3 k + 1 ⋮ ⋮ p⋮ ⋮ ⋮ ⋱ ��HHHp + 1k ` m ⋯

1 2 ` + 1 ⋯3 k + 1 ⋮ ⋮ ⋮⋮ ⋮ ⋮ ⋱ ⋮k ` m ⋯ p

p + 1

In the first picture, T ∣[p] is not a rectangle, so we may apply a negative rotationby the second case of Lemma 4.10, so consider the second picture. In thesecond picture, T ∣[p] is a rectangle and we know p + 1 exists in T since T ∣[p]

29

is an exceptional tableau for a rectangle shape. If p + 2 is in row k, column2, a (p, p + 1) rotation rule applies. If p + 2 is not in row k, column 2, then aB4-move applies.

Finally, consider the case ` = 2(k − 1) and T(k,1) = ` + 1. If T(k,2) ≠ ` + 2 andk > 3, then a B5-move applies. If T(k,2) = ` + 2 and k > 3, then the rotation(`, ` + 1) applies to T since ` − 1 is above `. If T(k,2) = ` + 2 and k = 3, then` = 4 = T(2,2) and T(3,1) = 5 so T contains

1 23 45

.

In this case, consider the subcases c = ab or c < ab with a = 2. If c = ab, thenT(1,3) = c + 1 since T /∈ E(λ). Either a B1-move applies if T(2,3) = c + 2 and a(c, c+1) rotation applies otherwise. On the other hand, if c < ab then a B2-ruleapplies. �

We may finally define the map ϕ from (15). The proof of Theorem 1.1from the introduction follows immediately from this definition and the last fewlemmas.

Definition 4.16. Given T ∈ SYT(λ) − E(λ), we define ϕ(T ) as follows. If1 ∈ Des(T ), define ϕ(T ) = (z, z − 1, . . . , i)T as in Corollary 4.9. If 1, 2 /∈ Des(T ),then Lemma 4.14 applies, so define ϕ(T ) using the specific B1, B2, B3 orrotation rule identified in the proof of that lemma. If 1 /∈ Des(T ) and 2 ∈ Des(T ),then Lemma 4.15 applies, so define ϕ(T ) using the specific B1, B2, B4, B5, ornegative rotation rule identified in the proof of that lemma. These rules coverall possible cases. By contruction, maj(ϕ(T )) = maj(T ) + 1.

We may define two poset structures on standard tableaux of a given shapeusing the preceding combinatorial operations. We call them “strong” and“weak” in analogy with the strong and weak Bruhat orders on permutations.Recall an inverse-transpose block rule is a block rule obtained from transposingthe diagrams in Figure 8 and reversing the arrows.

Definition 4.17. As sets, let P (λ) and Q(λ) be eitherSYT(λ) ∖ {minmaj(λ),maxmaj(λ)}

if λ is a rectangle with at least two rows and columns, or SYT(λ) otherwise.● (Strong SYT Poset) Let P (λ) be the partial order with covering rela-

tions given by rotations, block rules, and inverse-transpose block rulesincreasing maj by 1.

● (Weak SYT Poset) Let Q(λ) be the partial order with covering relationsgiven by S ≺ T if ϕ(S) = T or ϕ(T ′) = S′ where S′, T ′ are the transposeof S,T , respectively.


Corollary 4.18. As posets, P (λ) and Q(λ) are ranked with a unique minimaland maximal element. If λ is not a rectangle, the rank function is given byrk(T ) = maj(T ) − b(λ). If λ is a rectangle with at least two rows and columns,then the rank function is given by rk(T ) = maj(T ) − b(λ) − 2.Proof. By Corollary 4.2, P (λ) and Q(λ) have a single element of minimal majand of maximal maj. Any element T besides these is covered by ϕ(T ) andcovers ϕ(T ′)′, so is not maximal or minimal. By construction maj increases by1 under covering relations. The result follows. �

In Figure 9, we show an example of both the Weak SYT Poset and the StrongSYT poset for λ = (3,2,1). More examples of these partial orders are given athttps://sites.math.washington.edu/~billey/papers/syt.posets.

Remark 4.19. Observe that both the positive and negative rotation rulesapply equally well to any skew shape tableaux in SYT(λ/ν). The block rulesapply to skew shape tableaux as well when Tz is a straight shape tableau.However, in order to define the analogous posets on SYT(λ/µ), one mustinclude additional block moves. This is part of an ongoing project.

Remark 4.20. Lascoux–Schützenberger [LS81] defined an operation calledcyclage on semistandard tableaux, which decreases cocharge by 1. The cyclageposet on the set of semistandard tableaux arises from applying cyclage inall possible ways. Cyclage preserves the content, i.e. the number of 1’s, 2’s,etc. See also [SW00, Sect. 4.2]. Restricting to standard tableaux, cochargecoincides with maj, so the cyclage poset on SYT(n) is ranked by maj. However,cyclage does not necessarily preserve the shape, so it does not suffice to proveTheorem 1.1. For example, restricting the cyclage poset to SYT(32) gives aposet which has two connected components and is not ranked by maj, whileboth of our poset structures on SYT(32) are chains. A reviewer of [BKS20b]posed an interesting question: is there any relation between the cyclage posetcovering relations restricted to SYT(λ) and the two ranked poset structuresused to prove Theorem 1.1? We have not found one, but such a connectionwould be interesting if found.

5. Internal zeros for des on SYT(λ)

The results of Section 4 show that SYT(λ)maj(q) almost never has internalzeros. Adin–Elizalde–Roichman analogously considered the internal zeros ofthe descent number generating functions SYT(λ/ν)des(q) where des(T ) is thenumber of descents in a tableau T .

Question 5.1. [AER18, Problem 7.5] Is {des(T ) ∶ T ∈ SYT(λ/ν)} an intervalconsisting of consecutive integers, for any skew shape λ/ν? That is, doesSYT(λ/ν)des(q) ever have internal zeros?

31

Weak

(4 2 5 1 3 6)

(5 2 6 1 3 4) (4 3 5 1 2 6)

(5 3 6 1 2 4) (5 3 4 1 2 6)

(3 2 5 1 4 6)(6 3 4 1 2 5) (5 4 6 1 2 3)

(3 2 6 1 4 5) (5 2 4 1 3 6)(6 4 5 1 2 3)

(4 2 6 1 3 5)(6 2 4 1 3 5)

(4 3 6 1 2 5)(6 2 5 1 3 4)

(6 3 5 1 2 4)

Strong

(4 2 5 1 3 6)

(5 2 6 1 3 4) (4 3 5 1 2 6)

(5 3 6 1 2 4) (5 3 4 1 2 6)

(3 2 5 1 4 6)(6 3 4 1 2 5)(5 4 6 1 2 3)

(3 2 6 1 4 5)(5 2 4 1 3 6)(6 4 5 1 2 3)

(4 2 6 1 3 5)(6 2 4 1 3 5)

(6 2 5 1 3 4) (4 3 6 1 2 5)

(6 3 5 1 2 4)

Figure 9. Hasse diagram of the Weak SYT Poset and theStrong SYT Poset of λ = (3,2,1). Each tableau is representedby its row reading word in these pictures.

The minimum and maximum descent numbers are easily described as follows.The argument involves constructions similar to the sequences of vertical andhorizontal strips used in Definition 4.1.

Lemma 5.2. [AER18, Lemma 3.7(1)] Let λ/ν be a skew shape with n cells.Let c be the maximum length of a column and r be the maximum length of a


row. Then

min des(SYT(λ/ν)) = c − 1max des(SYT(λ/ν)) = (n − 1) − (r − 1) = n − r.

Indeed, it is easy to see that minmaj(λ) constructed as in Definition 4.1 hasλ′1 − 1 = c − 1 descents, and symmetrically that maxmaj(λ) has n − r descents.The arguments in Section 4 consequently resolve Question 5.1 affirmatively inthe straight-shape case.

Corollary 5.3. For λ ⊢ n, we have{des(T ) ∶ T ∈ SYT(λ)} = {λ′1 − 1, λ′1, . . . , n − λ1 − 1, n − λ1}.

In particular, SYT(λ)des(q) has no internal zeros.

Proof. First suppose λ is not a rectangle with at least two rows and columns.Iterating the ϕ map creates a chain from minmaj(λ) to maxmaj(λ). At eachstep, ϕ either applies a rotation rule or a block rule. Rotation rules preservedescent number. Block rules always increase the descent number by exactly 1.Since minmaj(λ) and maxmaj(λ) have the minimum and maximum numberof descents possible, the result follows.

If λ is a rectangle with at least two rows and columns, it is easy to see thatthe unique tableau of major index b(λ) + 2 has exactly one more descent thanminmaj(λ). The result follows as before by iterating the ϕ map. �

Remark 5.4. The same argument shows that SYT(λ)maj−des(q) also has nointernal zeros. Indeed, applying a rotation rule increases maj−des by 1 whilefixing des, and applying a B-rule fixes maj−des and increases des by 1. In thissense, the strong or weak posets P (λ) and Q(λ) have a Z×Z ranking given by(maj−des,des).

6. Internal zeros for fake degrees of Cm ≀ SnIn this section, we classify which irreducible representations appear in which

degrees of the corresponding coinvariant algebras for all finite groups of theform Cm ≀ Sn. The goal is to classify when the fake degrees bλ,k ≠ 0. We willuse the following helpful lemma which is straightforward to prove.

Lemma 6.1. Suppose that f and g are polynomials in Z[q] with non-negativecoefficients, that f has no internal zeros and has at least two non-zero coeffi-cients, and that g has no adjacent internal zeros. Then, fg has no internalzeros.

Lemma 6.2. Let λ = (λ(1), . . . , λ(m)) be a sequence of partitions. The polyno-mial SYT(λ)maj(q) has no internal zeros except when λ has a single non-empty

33

block λ(i) which is a rectangle with at least two rows and columns. In this lattercase, the only internal zero up to symmetry occurs at k = b(λ(i)) + 1.

Proof. If λ has only one nonempty partition, then the characterization ofinternal zeros follows from Theorem 1.1, so assume λ has two or more nonemptypartitions. From Theorem 2.15, we have

(17) SYT(λ)maj(q) = ( n∣λ(1)∣, . . . , ∣λ(m)∣

)q

m

∏i=1

SYT(λ(i))maj(q).

By MacMahon’s Theorem 2.5, we observe that the q-multinomial coefficientshave no internal zeros. Furthermore, the q-multinomial in (17) is not constantwhenever λ has two or more non-empty partitions. From Theorem 1.1, we knowSYT(λ(i))maj(q) has no adjacent internal zeros for any 1 ≤ i ≤m. Consequently,by Lemma 6.1, the overall product in (17) has no internal zeros. �

Theorem 6.3. Let λ be a sequence of m partitions with ∣λ∣ = n, and assumegλ(q) = ∑k bλ,kqk. Then for k ∈ Z, bλ,k ≠ 0 if and only if

k − b(α(λ))m

− b(λ) ∈⎧⎪⎪⎨⎪⎪⎩

0,1, . . . ,(n + 12

) −∑c∈λ

hc

⎫⎪⎪⎬⎪⎪⎭∖Dλ,

where Dλ is empty unless λ has a single non-empty partition λ(i) which is arectangle with at least two rows and columns, in which case

Dλ = {1,(n + 1

2) − ∑

c∈λ(i)hc − 1} .

Proof. By Theorem 2.31,

gλ(q) = qb(α(λ)) SYT(λ)maj(qm)

which implies bλ,k ≠ 0 only if k − b(α(λ)) is a multiple of m. By Lemma 6.2,we know SYT(λ)maj(q) has either no internal zeros or internal zeros at degree1 + b(λ) and degree one less than the maximal major index for λ in thecase of a rectangle with at least 2 rows and columns. By (8) and (9), theminimal major index for λ is b(λ) ∶= ∑i b(λ(i)), and the maximal major indexis b(λ) + (∣λ∣+12 ) −∑c∈λ hc. Hence, the result follows. �

Corollary 6.4. In type Bn, the irreducible representation indexed by (λ,µ)with ∣λ∣ + ∣µ∣ = n appears in degree k of the coinvariant algebra of G(2,1, n) ifand only if

k − ∣µ∣2

− b(λ) − b(µ) ∈ {0,1, . . . ,(n + 12

) −∑c∈λ

hc −∑c′∈µ

hc′} ∖D(λ,µ).


Example 6.5. Consider the type B6 case, where m = 2, d = 1, n = 6. Forλ = ((2), (31)), we have b(λ) = 1 and

gλ(q) = q26 + 2q24 + 4q22 + 5q20 + 7q18 + 7q16 + 7q14 + 5q12 + 4q10 + 2q8 + q6.

For µ = (∅, (33)), we have b(µ) = 3 and

gµ(q) = q24 + q20 + q18 + q16 + q12.

In both cases, the nonzero coefficients are determined by Corollary 6.4.

7. Deformed Gaussian Multinomial Coefficients

We now turn our attention to extending Theorem 6.3 to general Shephard–Todd groups G(m,d,n). We begin by introducing a deformation of the q-multinomial coefficients arising from Theorem 2.35 in the special case whenλ = ((α1), (α2), . . . , (αm)) is a sequence of one row partitions. After severallemmas, we give an alternative formulation for these deformed q-multinomialsin terms of inversion generating functions on words with a bounded first letter.

Definition 7.1. Let α = (α1, . . . , αm) ⊧ n be a weak composition of n with mparts. Recall the long cycle σm = (1,2, . . . ,m) ∈ Sm, so

σm ⋅ α = (αm, α1, α2, . . . , αm−1).

Let d ∣m, τ = σm/dm , and Cd = ⟨τ⟩ = ⟨σm/dm ⟩ so Cd acts on length m compositionsby (m/d)-fold cyclic rotations as in Definition 2.37. Set

[nα]q;d

∶= ∑σ∈Cdqb(σ⋅α)

[d]qnm/d(nα)qm

(18)

where

b(α) ∶=m

∑i=1

(i − 1)αi.

Note that when q = 1, we have [nα]1;d = (nα), and when d = 1, we have

[nα]q;1

= qb(α)(nα)qm , where m is the number of parts of α. As usual, we alsowrite [nk]q;d ∶= [

nk,n−k

]q;d

= [ nn−k,k]q;d, where m = 2 in this case. Note that [nα]q;d

is

invariant under the Cd-action on α, though this is not typically true of generalpermutations of α.

Example 7.2. Observe that (nα)qm alone is generally not divisible by [d]qnm/d .For example, if n = 5, α = (2,1,1,1), and d = 2, we have

( 52,1,1,1

)q4= q36 + 3q32 + 6q28 + 9q24 + 11q20 + 11q16 + 9q12 + 6q8 + 3q4 + 1

35

which is not divisible by [2]q5⋅4/2 = q10 + 1. However, ∑σ∈Cd qb(σ⋅α) = q8 + q6 and(q8 + q6)( 52,1,1,1)q4 is divisible by q

10 + 1 giving

[ 52,1,1,1

]q;2

= q34 + q32 + 3q30 + 3q28 + 6q26 + 5q24 + 8q22

+ 6q20 + 8q18 + 5q16 + 6q14 + 3q12 + 3q10 + q8 + q6.

See Figure 10 for a larger example.

100 200 300 400 500

2e6

4e6

6e6

8e6

1e7

1.2e7

Figure 10. A plot of the coefficients for the deformed q-multinomial [nα]q;d with α = (2,1,3,1,4,5) and d = 3.

Lemma 7.3. Given α = (α1, . . . , αm) ⊧ n, we have

b(σm ⋅ α) − b(α) = n −mαm,and

b(τ ⋅ α) − b(α) = nm/d −m(αm + αm−1 +⋯ + αm−m/d+1).

Proof. The second claim follows by iterating the first for τ = σm/dm . For the first,we have

b(σm ⋅ α) − b(α) = (α1 + 2α2 +⋯ + (m − 1)αm−1)− (α2 + 2α3 +⋯ + (m − 1)αm)

= α1 + α2 +⋯ + αm−1 − (m − 1)αm,

which simplifies to n −mαm. �

If α ⊧ n, let ↓iα be the vector obtained from α by decreasing αi by 1. Extendthe definition of (nα)q to m-tuples of integers by declaring (

nα)q= 0 if any αi

is negative. The following lemma is well known but we include a proof forcompleteness.


Lemma 7.4. We have the following recurrence for q-multinomial coefficients,

( nα1, . . . , αm

)q

=m

∑i=1

qα1+⋯+αi−1(n − 1↓iα

)q

.

Proof. By MacMahon’s Theorem, the left-hand side is the inversion numbergenerating function on length n words with αi copies of the letter i for each i.If the first letter in such a word is i, the number of inversions involving thefirst letter is α1 + α2 +⋯ + αi−1, from which the result quickly follows. �

The non-trivial deformation of the q-binomial coefficients in Definition 7.1has the following more explicit form. In particular, these rational functions arepolynomials with non-negative integer coefficients that satisfy a Pascal-typeformula.

Lemma 7.5. In the case d =m = 2, we have

(19) [nk]q;2

= qk + qn−k1 + qn

(nk)q2= qn−k(n − 1

k − 1)q2+ qk(n − 1

k)q2∈ Z≥0[q].

Proof. The first equality is immediate from Definition 7.1. For the second, weuse the well-known “q-Pascal” identities

(nk)q

= qk(n − 1k

)q

+ (n − 1k − 1

)q

= (n − 1k

)q

+ qn−k(n − 1k − 1

)q

,

which arise from Lemma 7.4. Thus,

qk(nk)q2= qk(n − 1

k)q2+ qn+n−k(n − 1

k − 1)q2

and

qn−k(nk)q2= qn+k(n − 1

k)q2+ qn−k(n − 1

k − 1)q2.

Hence,

(qk + qn−k)(nk)q2= (1 + qn)(qk(n − 1

k)q2+ qn−k(n − 1

k − 1)q2)

so the second equality in (19) holds. �

We next generalize Lemma 7.5 to all [nα]q;d for any α ⊧ n. The proof thatfollows is independent of Theorem 2.35, which can also be used to prove theyare polynomials with non-negative integer coefficients.

Theorem 7.6. Let α be a weak composition of n with m parts, and let d ∣m.Then

[nα]q;d

= ∑σ∈Cd

qb(σ⋅α)m/d

∑v=1

qm⋅((σ⋅α)1+⋯+(σ⋅α)v−1)( n − 1↓v (σ ⋅ α)

)qm.

37

In particular, [nα]q;d is a polynomial with non-negative coefficients.

Proof. Observe from the definition that (nα)q = (nσ⋅α

)q

for any σ ∈ Cd. Thus, byLemma 7.4, we can rewrite the numerator of [nα]q;d as

∑σ∈Cd

qb(σ⋅α)(nα)qm

=d

∑j=1

qb(τj ⋅α)( n

τ j ⋅ α)qm

=d

∑j=1

m

∑i=1

q�(i,j,α)( n − 1↓i (τ j ⋅ α)

)qm

where

(20) �(i, j, α) ∶= b(τ j ⋅ α) +m ⋅ ((τ j ⋅ α)1 +⋯ + (τ j ⋅ α)i−1).It is straightforward to check that ↓i (σm ⋅ α) = σm⋅ ↓i−1α, so that ↓i (τ j ⋅ α) =τ j ⋅ ↓i−jm/dα, where indices are taken modulo m. Thus,

∑σ∈Cd

qb(σ⋅α)(nα)qm

=m

∑i=1

d

∑j=1

q�(i,j,α)( n − 1↓i−jm/dα

)qm

.

Group the terms on the right according to the value i − jm/d ≡m t ∈ [m]. Notethat j ∈ [d] could be equivalently represented as j ∈ Z/d, though i ∈ [m] cannotbe treated similarly here. One may check that the set of (i, j) ∈ [m]×Z/d suchthat i − jm/d ≡m t can be described as

{(t + sm/d, s) ∶ s ∈ [−pt, d − 1 − pt]}where t = ptm/d+vt for some unique pt ∈ [0, d−1] and vt ∈ [m/d]. Consequently,

∑σ∈Cd

qb(σ⋅α)(nα)qm

=m

∑t=1

(d−1−pt

∑s=−pt

q�(t+sm/d,s,α))(n − 1↓tα

)qm.

Next, we evaluate the incremental change

�(t + (s + 1)m/d, s + 1, α) − �(t + sm/d, s,α)for a given s. Let β = τ s ⋅ α. By Lemma 7.3,

b(τ s+1 ⋅ α) − b(τ s ⋅ α) = b(τ ⋅ β) − b(β)= nm/d −m ⋅ (βm +⋯ + βm−m/d+1).

We also find

(τ ⋅ β)1+⋯ + (τ ⋅ β)t+(s+1)m/d−1 = βm−m/d+1 +⋯ + βm + β1 +⋯ + βt+sm/d−1so

(τ ⋅ β)1+⋯ + (τ ⋅ β)t+(s+1)m/d−1 − β1 −⋯ − βt+sm/d−1= βm−m/d+1 +⋯ + βm.


Combining these observations,

�(t + (s + 1)m/d, s + 1, α) − �(t + sm/d, s,α) = nm/d.It follows that

d−1−pt

∑s=−pt

q�(t+sm/d,s,α) = q�(t−ptm/d,−pt,α)[d]qnm/d

= q�(vt,−pt,α)[d]qnm/d .

Since we have a bijection [0, d − 1] × [m/d]→ [m] given by (p, v)↦ pm/d + v,we have

(21) ∑σ∈Cd

qb(σ⋅α)(nα)qm

= [d]qnm/dd−1

∑p=0

m/d

∑v=1

q�(v,−p,α)( n − 1↓v+pm/dα

)qm

,

proving the polynomiality of [nα]q;d.We can further refine (21). From (20), we observe that

�(v,−p,α) = �(v,0, τ−p ⋅ α),

and since τ = σm/dm , we have↓v+pm/dα = τ p⋅ ↓v (τ−p ⋅ α).

So,

q�(v,−p,α)( n − 1↓v+pm/dα

)qm

= q�(v,0,τ−p⋅α)( n − 1↓v (τ−p ⋅ α)

)qm,

which implies

∑σ∈Cd

qb(σ⋅α)(nα)qm

= [d]qnm/d ∑σ∈Cd

qb(σ⋅α)m/d

∑v=1

qm⋅((σ⋅α)1+⋯+(σ⋅α)v−1)( n − 1↓v (σ ⋅ α)

)qm.

The result follows by dividing by [d]qnm/d . �

In light of Theorem 7.6, we define the following polynomials.

Definition 7.7. Let α = (α1, . . . , αm) ⊧ n, and say 1 ≤ k ≤ m. Define theα, k-partial sum multinomial by

p(k)α (q) =

k

∑i=1

qα1+⋯+αi−1(n − 1↓iα

)q

.

Remark 7.8. By Lemma 7.4, p(m)α = (nα)q, and more generally the same

argument shows that

(22) p(k)α (q) = {w ∈ Wα ∶ w1 ≤ k}inv(q)

is an inversion number generating function.

39

It is very well-known that the multinomial coefficients can be written as aproduct of binomial coefficients. More generally, q-multinomial coefficients canbe written as a product of q-binomial coefficients. This holds true even for theα, k-partial sum multinomials as follows.

Lemma 7.9. Let α = (α1, . . . , αm) ⊧ n and 1 ≤ k ≤m. We have

(23) p(k)α (q) =

k

∏i=1

(α1 +⋯ + αiαi

)q

⋅m

∏i=k+1

(α1 +⋯ + αi − 1αi

)q

.

Proof. Recall that p(k)α (q) = {w ∈ Wα ∶ w1 ≤ k}inv(q). Partition the set {w ∈

Wα ∶ w1 ≤ k} into ( n−1α1+⋯+αk−1,αk+1,...,αm) subsets according to the placement of allk + 1, k + 2, . . . ,m’s in positions 2,3, . . . , n. For each such placement, there are(α1+⋯+αkα1,...,αk

) ways to place numbers 1,2, . . . , k in the remaining positions. Sinceeach inversion in a word w ∈ Wα is between two letters ≤ k, between two letters≥ k + 1, or between a letter ≤ k and a letter ≥ k + 1, it follows that

(24) {w ∈ Wα ∶ w1 ≤ k}inv(q) = (α1 +⋯ + αkα1, . . . , αk

)q

( n − 1α1 +⋯ + αk − 1, αk+1, . . . , αm

)q

by MacMahon’s Theorem 2.5. Factoring each q-multinomial in (24) into q-binomials gives (23). �

Corollary 7.10. Let α = (α1, . . . , αm) ⊧ n and 1 ≤ k ≤ m. The α, k-partialsum multinomial p

(k)α (q) is symmetric and unimodal.

Proof. A result of Andrews [And76, Thm. 3.9] states that the product ofsymmetric, unimodal polynomials with non-negative coefficients is symmetricand unimodal with non-negative coefficients. The q-binomials are symmetricwith non-negative coefficients, and it is a well-known, non-trivial fact that theyare also unimodal. See [Zei89] for a combinatorial proof of this fact and furtherhistorical references. The result now follows from Lemma 7.9. �

Lemma 7.11. Let α = (α1, . . . , αm) ⊧ n and 1 ≤ k ≤ m. Then p(k)α (q) ≠ 0 ifand only if α1 + ⋯ + αk > 0. In this case, p(k)α (q) has constant coefficient 1,degree Dα − αk+1 − ⋯ − αm where Dα = (n2) −∑ (

αi2) is the degree of (nα)q, and

has no internal zeros. Furthermore, p(k)α (q) is non-constant except when

● α1 +⋯ + αk = 0, in which case p(k)α (q) = 0;● α1 +⋯ + αk = 1 and αi = n − 1 for some i > k, in which case p(k)α (q) = 1;

or● αi = n for some i ≤ k, in which case p(k)α (q) = 1.

Proof. Each claim follows easily from the fact that p(k)α (q) is the inversion

generating function for {w ∈ Wα ∶ w1 ≤ k}. Alternatively, one may useLemma 7.9. �


We have the following summary statement.

Corollary 7.12. Let α be weak composition of n with m parts, and let d ∣m. Let α = ((α1), (α2), . . . , (αm)) be the corresponding sequence of one rowpartitions. Then,

[nα]q;d

= ∑σ∈Cdqb(σ⋅α)

[d]qnm/d(nα)qm

= ∑σ∈Cd

qb(σ⋅α)p(m/d)σ⋅α (qm) =

d

#{α}dg{α}

d(q).

Proof. The first equality is just the definition. The second equality followsfrom Theorem 7.6. The third equality follows from Theorem 2.35 and the factthat

∑σ∈Cd

qb(σ⋅α) = d#{α}d

({α}d)b○α (q).

�

Remark 7.13. We note that since [nα]q;d =d

#{α}dg{α}

d(q), we knew fromStembridge’s work that the deformed multinomial coefficients are polynomialsin q even though they are defined as rational functions. Our proof in The-orem 7.6 gives an alternate, direct proof of this fact without going throughrepresentation theory. Furthermore, we use the summation formula in Corollary7.12 to characterize the internal zeros of g{λ}

d(q) in the next section.

8. Internal zeros for G(m,d,n)

We can now extend the results of Section 6 to all Shephard–Todd groupsG(m,d,n). We thus give a remarkably simple and completely general de-scription for which irreducible representations appear in which degrees of thecoinvariant algebras of essentially arbitrary complex reflection groups. Recallthe notation established in Section 2.4. Let {λ}d be the orbit of λ under(m/d)-fold rotations in Cd = ⟨σm/dm ⟩.Definition 8.1. Given a sequence λ = (λ(1), . . . , λ(m)) with ∣λ∣ = n, let

α(λ) ∶= (∣λ(1)∣, . . . , ∣λ(m)∣) ⊧ n.Similarly, let α(λ) be the length m sequence of partitions whose ith partitionis the single row partition of size ∣λ(i)∣.

The map α may not be injective on {λ}d, though it has constant fiber sizessince α is Cd-equivariant. For example, when m = 4, d = 4, we have

α∶ ((2),∅, (12),∅)↦ ((2),∅, (2),∅)(∅, (2),∅, (12))↦ (∅, (2),∅, (2))((12),∅, (2),∅)↦ ((2),∅, (2),∅)(∅, (12),∅, (2))↦ (∅, (2),∅, (2)).

41

Generalizing Theorem 2.15, we have the following corollary of Stembridge’sTheorem 2.35 and Definition 7.1.

Corollary 8.2. Let λ be a sequence of m partitions with ∣λ∣ = n. Let {λ}d bethe orbit of λ under (m/d)-fold cyclic rotations. Then

g{λ}d(q) = #{λ}

d

d⋅ [ nα(λ)

]q;d

⋅m

∏i=1

SYT(λ(i))maj(qm).

Proof. By Theorem 2.35,

g{λ}d(q) = ({λ}

d)b○α (q)[d]qnm/d

SYT(λ)maj(qm).

We have

({λ}d)b○α (q) = #{λ}d

#{α(λ)}d⋅ ({α(λ)}d)b○α (q)

and

({α(λ)}d)b○α (q) = #{α(λ)}d

d∑σ∈Cd

qb(σ⋅α(λ)).

Consequently, using Theorem 2.15 and Definition 7.1, we have

g{λ}d(q) = #{λ}

d

d⋅ ∑σ∈Cd

qb(σ⋅α(λ))

[d]qnm/d⋅ ( nα(λ)

)qm⋅m

∏i=1

SYT(λ(i))maj(qm)

= #{λ}d

d⋅ [ nα(λ)

]q;d

⋅m

∏i=1

SYT(λ(i))maj(qm).

�

We will now prove the general classification theorem for nonzero fake degreesas mentioned in the introduction. The reader may find it useful to comparethe statement to the type A case in Theorem 1.1 and the Cm ≀ Sn case inTheorem 6.3.

Theorem 8.3. Let λ be a sequence of m partitions with ∣λ∣ = n ≥ 1, let d ∣m,and let {λ}d be the orbit of λ under the group Cd of (m/d)-fold cyclic rotations.Then b{λ}d,k ≠ 0 if and only if for some µ ∈ {λ} we have ∣µ(1)∣ +⋯ + ∣µ(m/d)∣ > 0and

k − b(α(µ))m

− b(µ) ∈⎧⎪⎪⎨⎪⎪⎩

0,1, . . . , ∣µ(1)∣ +⋯ + ∣µ(m/d)∣ + (n2) −∑

c∈µ

hc

⎫⎪⎪⎬⎪⎪⎭∖Dµ;d.

Here Dµ;d is empty unless either(1) µ has a partition µ of size n; or

(2) µ has a partition µ of size n − 1 and ∣µ(1)∣ +⋯ + ∣µ(m/d)∣ = 1,


where in both cases µ must be a rectangle with at least two rows and columns.In case (1), we have

Dµ;d ∶= {1,(n + 1

2) −∑

c∈µ

hc − 1} ,

and in case (2) we have

Dµ;d ∶= {1,(n

2) −∑

c∈µ

hc} .

Proof. Let α = α(λ). Using Corollary 8.2 and Corollary 7.12, we have

(25) g{λ}d(q) = #{λ}

d

d⋅ ∑σ∈Cd

qb(σ⋅α)p(m/d)σ⋅α (qm) ⋅

m

∏i=1

SYT(λ(i))maj(qm).

Thus, we consider the locations of the nonzero terms in

(26) p(m/d)σ⋅α (q)

m

∏i=1

SYT(λ(i))maj(q).

Recall that p(m/d)σ⋅α (q) = 0 whenever (σ ⋅ α)1 + ⋯ + (σ ⋅ α)m/d = 0, so assume

(σ ⋅α)1 +⋯+ (σ ⋅α)m/d > 0. Since SYT(λ(i))maj(q

TABLEAU POSETS AND THE FAKE DEGREES OFkonvalinka/zeros.pdf · The major index statistic on permutations was introduced by Percy MacMahon in his seminal works [Mac13, Mac17]. At rst

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